DFT Linearity

Explore the linearity property of the Fourier transform.

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Linearity implies that the DFT of the sum of input signals is the sum of the DFTs of the individual input signals. Let’s find out how.


Assume that two input signals x1[n]x_1[n] and x2[n]x_2[n] have DFTs given by X1[k]X_1[k] and X2[k]X_2[k], respectively. If that’s the case, the DFT of their sum is equal to the sum of their individual DFTs.

DFT x1[n]X1[k]DFT x2[n]X2[k]\begin{align*} \text{DFT }x_1[n]\quad \rightarrow\quad X_1[k]\\ \text{DFT }x_2[n]\quad \rightarrow\quad X_2[k] \end{align*}

This implies that:

DFT {y[n]=x1[n]+x2[n]}Y[k]=X1[k]+X2[k]\text{DFT }\Big\{y[n]=x_1[n]+x_2[n]\Big\}\quad \rightarrow\quad Y[k]=X_1[k]+X_2[k]

This is because:

Y[k]=n=0N1{x1[n]+x2[n]}ej2πkNn=n=0N1x1[n]ej2πkNn+n=0N1x2[n]ej2πkNn=X1[k]+X2[k]\begin{align*} Y[k]&=\sum_{n=0}^{N-1}\Big\{x_1[n]+x_2[n]\Big\}e^{-j2\pi\frac{k}{N}n}\\ &=\sum_{n=0}^{N-1}x_1[n]e^{-j2\pi\frac{k}{N}n}+\sum_{n=0}^{N-1}x_2[n]e^{-j2\pi\frac{k}{N}n}\\ &=X_1[k]+X_2[k] \end{align*}

This helps determine the DFT of a signal without explicit calculations.


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